Populating Next Right Pointers in Each Node II

题目地址： https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

 

关键思路：讲节点的左右子节点链接起来，遍历节点的next节点，即可。

class Solution {

public:

    TreeLinkNode* gethead(TreeLinkNode* root)

    {

        if(root == NULL)

            return NULL;

        if(root->left != NULL)

            return root->left;

        else

            return root->right;

    }

    

    TreeLinkNode* gettail(TreeLinkNode* root)

    {

        if(root == NULL)

            return NULL;

        if(root->right != NULL)

            return root->right;

        else

            return root->left;

    }



    void connect(TreeLinkNode *root) 

    {    

        TreeLinkNode* head = NULL;

        TreeLinkNode* tail = NULL;

        if(root == NULL)

            return;



        TreeLinkNode* pcur = root;

        while(root != NULL)

        {

            TreeLinkNode* headtmp = gethead(root);

            TreeLinkNode* tailtmp = gettail(root);



            if(headtmp == NULL)

            {

                root= root->next;

                continue;

            }



            if(headtmp != tailtmp)

                headtmp->next = tailtmp;



            if(tail != NULL)

            {

                tail->next = headtmp;

                tail = tailtmp;

            }

            else

            {

                head = headtmp;

                tail = tailtmp;

            }

            root = root->next;

        }



        while(pcur != NULL)

        {

            TreeLinkNode* tmp = gethead(pcur);

            if(tmp != NULL)

            {

                connect(tmp);

                break;

            }

            pcur = pcur->next;

        }

    }



};